Problema 2A22

Base de Cálculo: 1 mol de $\ce{X2Y}$ Pela estequiometria do composto: $$\frac{n_{\ce{X}}}{2}=\frac{n_{\ce{Y}}}{1}=\frac{n_\ce{X2Y}}{1}$$ $$n_\ce{X}=\pu{2 mol}$$ $$n_\ce{Y}=\pu{1 mol}$$ Relacionando as massas molares a partir da fração mássica de $\ce{X}$: $$f_\ce{X}=\frac{m_{\ce{X}}}{m _\text{total}}$$ $$0,2=\frac{2\cdot M_{X}}{2\cdot M_{X} + M_\ce{Y}}$$ $$8\cdot M_{\ce{X}}=\pu{M_\ce{Y}}$$ Base de Cálculo: 1 mol de $\ce{XY2}$ : Pela estequiometria do composto:

$$\frac{n_{\ce{X}}}{1}=\frac{n_{\ce{Y}}}{2}=\frac{n_\ce{XY2}}{1}$$ $$n_\ce{X}=\pu{2 mol}$$ $$n_\ce{Y}=\pu{1 mol}$$ Cálculo da fração mássica de Y nesse composto: $$f_\ce{Y}=\frac{m_{\ce{Y}}}{m _\text{total}}$$ $$f_{\ce{Y}}=\frac{2\cdot M_\ce{Y}}{M_{\ce{X}}+ 2\cdot M_\ce{Y}}$$ $$f_\ce{Y}=\frac{2\cdot8\cdot M_\ce{X}}{M_\ce{X}+2\cdot8\cdot M_\ce{X}}$$ $$\boxed{f_\ce{Y}=94\%}$$